Given the following matrix equation, $$\begin{bmatrix}a && b \\ c&& d\end{bmatrix}^n\begin{bmatrix}\alpha\\\beta\end{bmatrix}=\begin{bmatrix}\gamma \\ \kappa\end{bmatrix}$$
$\alpha, \beta, \gamma, \kappa$ and $n$ are all known quantities and $n\subset\mathbb{N}$ and $\{\alpha, \beta, \gamma, \kappa, a, b, c, d\} \subset\mathbb{C}$ . Can we find the values of $a, b, c$ and $d$?
On
Assume that we are over $\mathbb{C}$. One has $2$ algebraic equations linking the $4$ unknowns $a,b,c,d$; then, in general, there are many solutions that depend on $2$ parameters. Thus, a better question is: "can you obtain at least one particular solution ?
The answer is yes when $\alpha \beta\not= 0$. Take $b=c=0$ and $a,d$ s.t; $a^n=\gamma/\alpha,d^n=\kappa/\beta$.
You can study by yourself the other cases; to work (even a little) is good for health.
We are given an initial state $\mathrm x_0 \in \mathbb R^2$ and a final state $\mathrm x_f \in \mathbb R^2$. We would like to find $\mathrm A \in \mathbb R^{2 \times 2}$ such that $\mathrm A^n \mathrm x_0 = \mathrm x_f$ for some $n \in \mathbb N$. Let $\Phi_n := \mathrm A^n$. We then have a linear system in $\Phi_n$
$$\Phi_n \mathrm x_0 = \mathrm x_f$$
Vectorizing, we obtain an underdetermined linear system in a more standard form
$$(\mathrm x_0^T \otimes \mathrm I_2) \, \mbox{vec} (\Phi_n) = \mathrm x_f$$
Once we have found a solution $\Phi_n$, we can try to find $\mathrm A \in \mathbb R^{2 \times 2}$ and $n \in \mathbb N$ such that $\mathrm A^n = \Phi_n$.
If $\mathrm A$ were given and rational, and if the initial and final states were also rational, then we would only have to determine $n$, which is the orbit problem.